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I'm presently working in a non-ZF set theory, where there are proper classes. (Think MK or VNBG.) And I'm interested in how to think about the possibility (or impossibility) of proper classes with cardinalities which are in some sense different. (So in this set theory we drop the von Neumann assumption, on which every proper class has the same cardinality.) Can someone point me to resources where people talk about this? I'm familiar with at least a slice of the relevant standard resources in math and phil math - MK's discussion, Parsons' discussion, Maddy's, etc. - but I've been having trouble tracking down discussions of this particular question.

Gratefully in advance,

Anon

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    $\begingroup$ One minor note: the language of “size”, “larger” etc for cardinality/equinumerosity can be quite misleading in settings without choice. Many of the results that make cardinality fit the intuition for a notion of “size” — most importantly, the duality between “injects into” and “surjects onto”; also cardinal trichotomy — depend on choice and may fail in its absence. So if you don’t want to presuppose choice, it’s good to at least be careful with that language. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Mar 13 at 11:41

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The assertion that all proper classes are equinumerous is equivalent over GBc class theory to the axiom of global choice, which is the assertion that there is a global well-ordering of the universe of sets.

Namely, if all proper classes are equinumerous (that is, have the same size), then in particular, the class of all sets $V$ is equimumerous with Ord, the class of ordinals, and this implies global choice. Conversely, if global choice holds, then all classes are equinumerous with Ord. I wrote a blog post, The global choice principle in Gödel-Bernays set theory explaining several further equivalent formulations of this principle.

The assertion that all proper classes are equinumerous is also known as the axiom Limitation of Size. Some people take this as one of the axioms of Gödel-Bernays set theory. I have criticized this axiom, however, on the grounds that we seem to have no convincing justification for it, specifically, that stands apart from the justifications that we might have for the replacement axiom and the global choice principle. Basically, the so-called Limitation of Size axiom is simply a method of combining two axioms into one statement. But I see no reason or advantage in doing that, and so I prefer to consider the replacement axiom and global choice axioms separately.

Meanwhile, one might mention that global choice cannot be proved in the theory GBc, which is Gödel-Bernays set theory but only with the axiom of choice for sets and not global choice. For example, there is a model of ZFC set theory in which the universe cannot be linearly ordered. By taking definable classes, one gets a model of GBc without global choice.

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  • $\begingroup$ This is, of course, incredibly helpful. So (just to restate one of your points, to make sure I'm tracking): if I were working in GBc, and I wanted to think about proper classes w different cardinalities, then I would just drop the axiom of global choice and think about how to understand the different sizes of proper classes in that context? $\endgroup$
    – Anonymous grad student
    Commented Mar 12 at 23:52
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    $\begingroup$ Yes, that's right. If you want classes to exhibit different sizes, you need to drop global choice, but then it is possible. There will be a largest size, the equinumerosity class of $V$, and other sizes, such as the equinumerosity class of Ord. The class of surreal numbers is equinumerous with the class of all ordinal binary sequences $2^{<\text{Ord}}$. (By the way, if you find a post helpful, you can vote it up by clicking the uparrow button.) $\endgroup$
    – Joel David Hamkins
    Commented Mar 12 at 23:58
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    $\begingroup$ @MikeBattaglia The axiom V=HOD is equivalent over ZFC to the existence of a first-order definable (without parameters) global well-ordering of the universe. $\endgroup$
    – Joel David Hamkins
    Commented Mar 13 at 6:44
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    $\begingroup$ @Anonymousgradstudent Since every class is a subclass of $V$, this will necessarily be the largest class. $\endgroup$
    – Joel David Hamkins
    Commented Mar 13 at 6:44
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    $\begingroup$ Also related are the following: mathoverflow.net/questions/124494/… AND mathoverflow.net/questions/280526/… $\endgroup$
    – Ali Enayat
    Commented Mar 13 at 11:22

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